How do we know that the quotient group $\bar{E} = E/Z(E)$ is an elementary abelian group

Question

My question is:

How do we know that the quotient group $\bar{E}=E/Z(E)$ is an elementary abelian group?

Please find below some background information on the different relevant groups involved from the paper "Quantum Error Correction via Codes over $GF(4)$"

The Error Group $E$

Say I have a group $E$ of tensor products $\pm w_{1} \otimes \dots \otimes w_{n}$ and $\pm i w_{1} \otimes \dots \otimes w_{n}$.

This group $E$ is the error group, where each $w_{j}$ is an element of the Pauli Group {$I, X, Y, Z$}. $E$ is a subgroup of the unitary group $U(2^{n})$. As the elements in $E$ consist of tensor product on $n$ elements and each of those $n$ elements can be one of $4$ possible operators, this means that there are $4^{n}$ possible elements of $E$, before considering phase. Also, each element can have $4$ possible phases. Meaning that the order of $E$ is $$|E|=4 \times 4^{n}=2^{2n+2}$$

The Centre of $E$, $Z(E)$

• $Z(E)$ denotes the centre of $E$.
• $Z(E)=$ {$\pm I, \pm iI$}.
• $|Z(E)|=4$

The Quotient Group $\bar{E}$

$\bar{E}$ denotes a $2n$-dimensional binary vector space. It's elements are written $(a|b)$, equipped with inner product $$((a|b),(a'|b')) = a.b' + a'b$$ and this is a symplectic inner product as $((a|b),(a|b))=0$.

This $\bar{E}$ is actually the quotient group $\bar{E} = E/Z(E)$

As such, $|\bar{E}|= \frac{2^{2n+2}}{4} = 2^{2n}$ (This makes sense as we had previously established that it was a $2n$-dimensional binary vector space).

However, the paper then asserts that $\bar{E}$ is an elementary abelian group.

Attempts to understand and questions

In my attempts to understand the above statement I found on Wikipedia "an elementary abelian group is an abelian group in which elements other than the identity have the same order".

I was trying to make sense of this by thinking what the elements of $\bar{E}$ might look like. $\bar{E}$ consists of the left cosets of $Z(E)$ in $E$. So $\bar{E}$ might look something like:

$\bar{E} =${$e_{1}\langle iI\rangle, \dots, e_{2^{2n+2}}\langle iI \rangle$}

where $\langle iI \rangle=Z(E)$ and the operation above is matrix multiplication.

Subquestions:

1. Is this a correct interpretation of what elements in $\bar{E}$ look like?
2. If $1$ is correct, how does every element of the above form have the same order (except for the identity)
3. Is our reason for knowing $\bar{E}$ is abelian because its group operation is the symplectic inner product $$((a|b),(a'|b'))= a.b' + a'.b = a'.b + a.b'=((a'|b'),(a|b)),$$ meaning all elements of $\bar{E}$ commute?

Answer 1

First, the answers to the subquestions:

1. Technically, yes, but this is unhelpful.
2. See below.
3. No, the group structure has nothing to do with the inner product.

Now, let's describe how the multiplication works in $\bar{E}$. I suspect this needs discussion, because otherwise you wouldn't mention the inner product.

In fact, first let's describe how multiplication works in $E$. Consider $e_1=kw_1\otimes\ldots\otimes w_n$ and $e_2=\ell u_1\otimes\ldots\otimes u_n$, where $k,\ell\in\{\pm1,\pm i\}$. Then $$ e_1e_2 = (k\ell)(w_1u_1)\otimes\ldots\otimes(w_nu_n). $$ (This is how tensor products of linear operators work.) Every $w_ku_k$ is a product of two Pauli matrices, so it can always be represented as some Pauli matrix times a scalar from $\{\pm1,\pm i\}$. By the way, because of that scalar factor, your "Pauli Group" $\{I,X,Y,Z\}$ is not really a group. (Make sure you understand the previous sentence! If not, ask another question about that. Note that this is a statement just about Pauli matrices, not about their tensor products which are in $E$.) So, to represent the result in the usual form "scalar times the tensor product of Pauli matrices", collect all the scalars into a single factor in front of everything.

Example. Let $n=3$, $e_1=X\otimes Y\otimes Z$, and $e_2=iI\otimes X \otimes Z$. Then $e_1e_2 = i(XI)\otimes(YX)\otimes(ZZ) $ $= iX\otimes(-iZ)\otimes I = (i\cdot(-i))X\otimes Z\otimes I = X\otimes Z\otimes I$. Make sure you understand every step of this calculation and can find a product of, say, $-Z\otimes Y\otimes X$ and $-X\otimes Y\otimes Z$!

At this point stop and think what is the identity element in $E$.

Now, what is an element of the quotient group $\bar{E}$? Sure, it is a coset. By the way, in our case it doesn't matter whether it is a right or left coset, because the elements of the centre is, by the centre definition, ... Complete this statement and make sure you understand why this explains that it doesn't matter!

What do the cosets look like? Consider some $e_1\in E$ introduced above. The corresponding element of $\bar{E}$ is the coset $e_1 \langle iI\rangle = \{e_1,-e_1,ie_1,-ie_1\}$ (let's call it $\bar{e}_1$; pay attention to overlines in the rest of the answer!). The four coset representatives differ by the scalar prefactor only.

How do we do multiplication in $\bar{E}$? Say, if $\bar{e}_1,\bar{e}_2\in \bar{E}$, how do we find $\bar{e}_1\bar{e}_2$? As always in the quotient groups -- take any representatives $e_1\in\bar{e}_1$ and $e_2\in\bar{e}_2$, find their product in $E$, and then take the coset of the result: $\{e_1e_2,-e_1e_2,ie_1e_2,-ie_1e_2\}$. Make sure you understand why the resulting coset doesn't depend on your choice of $e_1,e_2$ from the given $\bar{e}_1,\bar{e}_2$!

What is the identity element in $\bar{E}$?

Now we need to show that $\bar{e}_1\bar{e}_2 = \bar{e}_2\bar{e}_1$. For that, note that for any $A,B\in\{I,X,Y,Z\}$, we have either $AB=BA$ or $AB=-BA$. Check that this is true for all $10$ combinations of $A,B$! Using this, we see that for any representatives $e_1,e_2$, when we calculate $e_1e_2$ and $e_2e_1$, the only difference that arises is an additional factor $-1$ in some products $w_ku_k$. These factors are all collected together as a product in the scalar prefactor, resulting in either $e_2e_1=e_1e_2$ or $e_2e_1=-e_1e_2$. (Make sure you understand why this works! Check this on the specific example of $e_1$ and $e_2$ in the example above.) However, $e_1e_2$ and $-e_1e_2$ fall into the same coset. Thus $\bar{E}$ is abelian.

To show that every non-identity element has the same order, first find $II$, $XX$, $YY$, and $ZZ$ (in the form "scalar times a Pauli matrix"), and then it should become clear that for any $\bar{e}\in\bar{E}$ the product of a certain number of instances of $\bar{e}$ is always equal to the identity element. If it's not clear, try taking the coset of $e_1$ from the example above and multiply it by itself until you get the identity element (note: in $\bar{E}$, not in $E$!).